JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 27)
White light is passed through a double slit and interference is observed on a screen 1.5 m away. The separation between the slits is 0.3 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringes. the difference in wavelengths of red and violet light is ................ nm.
Answer
300
Explanation
Position of bright fringe y = n$${{D\lambda } \over d}$$
y1 of red = $${{D{\lambda _r}} \over d}$$ = 3.5 mm
$$\lambda$$r = 3.5 $$\times$$ 10$$-$$3 $${d \over D}$$
Similarly, $$\lambda$$v = 2 $$\times$$ 10$$-$$3 $${d \over D}$$
$${\lambda _r} - {\lambda _v} = (1.5 \times {10^{ - 3}})\left( {{{0.3 \times {{10}^{ - 3}}} \over {1.5}}} \right)$$
= 3 $$\times$$ 10$$-$$7 = 300 nm
y1 of red = $${{D{\lambda _r}} \over d}$$ = 3.5 mm
$$\lambda$$r = 3.5 $$\times$$ 10$$-$$3 $${d \over D}$$
Similarly, $$\lambda$$v = 2 $$\times$$ 10$$-$$3 $${d \over D}$$
$${\lambda _r} - {\lambda _v} = (1.5 \times {10^{ - 3}})\left( {{{0.3 \times {{10}^{ - 3}}} \over {1.5}}} \right)$$
= 3 $$\times$$ 10$$-$$7 = 300 nm
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